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(D) For a probability distribution,the sum of all probabilities must be equal to $1$:$\sum P(X=x) = 0 + k + 2k + 5k^2 + 2k^2 + 3k = 1$$7k^2 + 6k - 1 =

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$\frac{326}{49}$

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(D) For a probability distribution,the sum of all probabilities must be equal to $1$:$\sum P(X=x) = 0 + k + 2k + 5k^2 + 2k^2 + 3k = 1$$7k^2 + 6k - 1 = 0$$7k^2 + 7k - k - 1 = 0$$7k(k + 1) - 1(k + 1) = 0$$(k + 1)(7k - 1) = 0$Since $k$ must be positive for probabilities,$k = \frac{1}{7}$.The mean $E(X)$ is given by $\sum x_i P(X=x_i)$:$E(X) = (0 	imes 0) + (2 	imes k) + (4 	imes 2k) + (6 	imes 5k^2) + (8 	imes 2k^2) + (10 	imes 3k)$$E(X) = 0 + 2k + 8k + 30k^2 + 16k^2 + 30k$$E(X) = 46k^2 + 40k$Substituting $k = \frac{1}{7}$:$E(X) = 46(\frac{1}{7})^2 + 40(\frac{1}{7})$$E(X) = \frac{46}{49} + \frac{40}{7} = \frac{46 + 280}{49} = \frac{326}{49}$Thus,the correct option is $(d)$.

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

If the probability distribution of a random variable $X$ is given by the following table,then the mean of $X$ is:
$X = x$ $0$ $2$ $4$ $6$ $8$ $10$
$P(X = x)$ $0$ $k$ $2k$ $5k^2$ $2k^2$ $3k$

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| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
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If the probability distribution of a random variable $X$ is given by the following table,then the mean of $X$ is:$X = x$$0$$2$$4$$6$$8$$10$$P(X = x)$$0$$k$$2k$$5k^2$$2k^2$$3k$